Find $n$ for the most stable structure of $\mathrm{ClF_3}$

• In $\mathrm{ClF_3}$, central atom $\mathrm{Cl}$ has 5 electron pairs around it:
  • 3 bond pairs
  • 2 lone pairs
• Thus, the electron-pair geometry is trigonal bipyramidal.
• In a trigonal bipyramidal arrangement, lone pairs prefer equatorial positions because that minimizes repulsion.
• Therefore, in the most stable structure of $\mathrm{ClF_3}$, both lone pairs occupy equatorial positions.

Hence, $n = 2$

Now check which ions have $n=2$ unpaired electrons

We determine the electronic configuration of each ion.

A. $\mathrm{V^{3+}}$

• Atomic number of V = 23
• Neutral V: $[\mathrm{Ar}]\,3d^3 4s^2$
• Remove 2 electrons from $4s$ and 1 from $3d$: $\mathrm{V^{3+}} = [\mathrm{Ar}]\,3d^2$
• Number of unpaired electrons in $3d^2$ = 2

So, A is correct.

B. $\mathrm{Ti^{3+}}$

• Atomic number of Ti = 22
• Neutral Ti: $[\mathrm{Ar}]\,3d^2 4s^2$
• Remove 2 electrons from $4s$ and 1 from $3d$: $\mathrm{Ti^{3+}} = [\mathrm{Ar}]\,3d^1$
• Number of unpaired electrons = 1

So, B is incorrect.

C. $\mathrm{Cu^{2+}}$

• Atomic number of Cu = 29
• Neutral Cu: $[\mathrm{Ar}]\,3d^{10}4s^1$
• Remove $4s$ electron first, then one from $3d$: $\mathrm{Cu^{2+}} = [\mathrm{Ar}]\,3d^9$
• Number of unpaired electrons in $3d^9$ = 1

So, C is incorrect.

D. $\mathrm{Ni^{2+}}$

• Atomic number of Ni = 28
• Neutral Ni: $[\mathrm{Ar}]\,3d^8 4s^2$
• Remove 2 electrons from $4s$: $\mathrm{Ni^{2+}} = [\mathrm{Ar}]\,3d^8$
• Number of unpaired electrons in $3d^8$ = 2

So, D is correct.

E. $\mathrm{Ti^{2+}}$

• Neutral Ti: $[\mathrm{Ar}]\,3d^2 4s^2$
• Remove 2 electrons from $4s$: $\mathrm{Ti^{2+}} = [\mathrm{Ar}]\,3d^2$
• Number of unpaired electrons = 2

So, E is correct.

Select the correct option

The ions having 2 unpaired electrons are: $\mathrm{V^{3+}},\ \mathrm{Ni^{2+}},\ \mathrm{Ti^{2+}}$ i.e. A, D and E only.

Therefore, the correct option is: $\boxed{\text{B}}$

Comparison with stored correct answer

• Derived answer: B
• Stored correct answer: B

They agree.